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Surreal numbers

  1. May 11, 2008 #1
    How do you show that the surreal numbers form a proper class?
     
  2. jcsd
  3. May 12, 2008 #2

    matt grime

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    I suspect it goes along the lines of: show that there is proper class of ordinals, and show that for each ordinal one can construct inequivalent surreal numbers, or something.

    Since there are 'more' ordinals than cardinals, and there is a proper class of cardinals (if there is a set of cardinals, C, what is the cardinality of the power set of C?), this becomes easy - if I've understood the one thing I read about surreal numbers and some link to ordinals, and that;s a big if.
     
  4. May 12, 2008 #3
    I don't understand why the surreal numbers don't form a complete ordered field, in which case the real numbers wouldn't be the only complete ordered field. I guess technically if the surreals form a proper class then it wouldn't count, but that's... wrong, somehow.
     
  5. May 12, 2008 #4
    Never mind, I was reading the Wiki article, and I apparently misunderstood some things.
     
    Last edited: May 12, 2008
  6. May 12, 2008 #5

    matt grime

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    I don't know what you've read so I can't really comment, but completeness usually requires some notion of a metric - i.e. a mapping into R, the real numbers. So what is the norm of any non-real surreal?
     
  7. May 12, 2008 #6
    If a metric needs to be real-valued, and if the surreals form a proper class (or indeed a set bigger than reals) then it can't be defined.

    But I haven't read anything on the "norm" of surreal numbers; maybe it's defined differently.
     
  8. May 12, 2008 #7

    matt grime

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    It does.



    they do - apparently.

    Checking the definitions is always a good idea.
     
  9. May 12, 2008 #8
    I guess that answers the question about complete ordered fields.
     
  10. May 12, 2008 #9
    (Rehash of my earlier post... only this time, correct)
    Ignoring that the surreals form a proper class rather than a set, you'd need to prove that you can't define a metric on them that turns them into a complete ordered field.

    One way to see that the surreal numbers are not a complete ordered field is to use the axioms of a complete ordered field to show that they satisfy the archimedean property, which says that there are no infinitesimals other than 0 and no infinities. Since the surreal numbers have non-zero infinitesimals and infinities, they cannot be a complete ordered field.
     
    Last edited: May 12, 2008
  11. May 14, 2008 #10

    Hurkyl

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    The easiest way to show the surreals are a proper class is by contradiction. Assume they form a set, then construct a new surreal number not in that set.

    (hint... can you find one bigger than everything in that set?)
     
  12. May 14, 2008 #11
    If X is the set of all surreal numbers, then so is [tex]\{ X|\emptyset\}[/tex]. But this only leads to a contradiction assuming well-foundedness.
     
  13. May 14, 2008 #12

    Hurkyl

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    Or by trichotomy....
     
  14. May 15, 2008 #13
    How does it violate trichotomy?
     
  15. May 15, 2008 #14

    Hurkyl

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    You've asserted:

    X is the set of all surreal numbers
    {X | } is a surreal number

    and it's a basic fact of surreal numbers that

    {X | } is larger than every member of X

    so....


    Now that I think of it, I'm pretty sure it's also a theorem that every surreal number is a well-founded set.
     
    Last edited: May 15, 2008
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